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University  of  California. 


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THE 


SECRET  OF  THE  CIRCLE 


AND 


T 


HE   SQUARE 


BY 


J.  C.  WILLMON 


^LIF 
AUTHOR'S  EDITION 


McBRIDE  PRESS 
316  W.  Second  Street.  Los  Angeles,  Cal. 


COPYRIGHT   BY 

JEREMY  C.  WILLMON 
1905 


OF   TUt 

OF 


PREFACE. 


It  is  my  intention  to  demonstrate  the  possibility  of 
constructing  a  straight  line  equal  to  any  given  arc  of  a 
circle,  and  through  this  problem  to  construct  a  square, 
equal  in  area  to  any  circle  and  a  circle  equal  in  area  to 
any  square,  with  solutions  of  kindred  geometrical  pro- 
blems. Many  other  geometrical  problems  are  possible 
through  these  demonstrations. 

Another  problem  that  may  prove  interesting,  is 
the  division  of  angles  into  any  number  of  equal  parts; 
this  problem  comes  under  the  domain  of  the  higher 
plane  curves 

Through  the  problem  of  constructing  a  line  equal 
to  the  square  of  any  line,  it  becomes  possible  to  con- 
struct a  line  equal  to  any  line  multiplied  by  any  other 
line,  and  through  the  two  problems  it  becomes  possible 
to  construct  a  line  equal  to  the  cube  of  any  line. 


In  Figure  i  (which  is  not  referred  to  in  any  of  the 
problems  and  proofs,  but  is  included  herein  that  the 
reader  may  gain  a  clearer  idea  of  the  Secret  of  the  Circle 
and  of  the  figures  and  problems  following)  A  C  is  the 
diameter,  and  A  J  is  tangent  to  the  circle  at  A;  the  first 
lines  bisecting  the  two  semi-circles  equal  the  perimeter 
of  a  square  inscribed  in  the  circle;  the  second  bisectors 
equal  a  polygon  of  eight  sides  inscribed  in  the  circle; 
the  third  a  polygon  of  sixteen  sides;  the  fourth  a  poly- 
gon of  thirty-two  sides;  the  fifth  a  polygon  of  64  sides; 
the  sixth  a  polygon  of  12S  sides;  the  seventh  a  polygon 
of  256  sides;  the  eighth  a  polygon  of  512  sides;  the 
ninth  a  polygon  of  1024  sides;  the  tenth  a  polygon  of 
2048  sides  and  so  on;  when  the  bisections  and  right 
angles  approach  and  coincide  with  the  tangent,  the  line 
J  A  J  equals  the  circumference  of  the  circle.  The  area 
ot  a  circle  equals  radius  X  Yi  circumference;  then  rec- 
tangle a  e  f  j  =  area  of  the  circle.  If  a  d  be  made  % 
circumference  and  d  b  be  made  j^  a  c  cutting  the  circle 
at  b,  then  a  j  X  a  e  =  a  d  X  a  c  =  a  b'^;  then  a  b  =  one 
side  of  a  square  containing  the  same  area  as  the  circle. 

J.  C.  WILI.MON, 
Los  Angeles,  Cal.,   December  1904. 

3 


Figure  I 


Figure  ii 


5> 


Figure  III 


11 


Figure  IV 


13 


FIGURE  V 


15 


v 


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Figure  vi 


17 


) 


Figure  vii 


19 


PROBLEM  I 

To  find  a  line  which  shall  equal  an}^  giv^en  arc. 
Let  the  giv^en  arc  be  a  k  f ,  Figure  2,  draw  chord 
a  f;  at  a  erect  the  tangent  a  j  _\_  to  a  c  the  diameter  of 
circle  a  k  f  b;  draw  a  k  s  the  bisector  of  <  j  a  f;  draw 
f  s  j_  to  a  f;  draw  a  t  the  bisector  of  <  j  a  s;  draw  s  t 
_L  to  a  s;  and  so  on  indefinitely,  approaching  the  limit 
a  j ;  then  a  j  is  the  required  line. 

PROOF 

Let  ebe  the  center  of  the  circle. 

Bisect  <  a  e  f  at  k;  then  bisect  ■<  a  e  k  at  1,  etc. 

(Ref  I.     An  <  at  the  center  is  measured  by  }4 
the  intercepted  arc.) 

The  bisector  a  k  s  of  <  j  a  f ,  bisects  the  arc  a  k  f 
at  k. 

The  bisector  a  t  of  <  j  as,  bisects  the  area  k  at  1. 

(Ref  IL     An  <  formed  by  a  tangent  and  a  chord 
is  measured  by  j4  the  intercepted  arc.) 

The  bisectors  of  <s  formed  by  the  bisectors  of  <s 
jaf;jas;jat,    etc.,    and  the  bisectors  of    <s  a  e  f ; 

21 


a  ek,  a  el,  etc.,  being  measured  by  the  same  arcs,  meet 
in  a  point  in  the  circumference  of  the  circle  at  the 
points k,  1,  o,  etc. 

(leisj_aks,  Ref.  III.  The  _\_  erected  at  the  mid- 
dle of  a  chord  passes  through  the  center  of  the  circle 
and  bisects  the  arc  of  the  chord.) 

f  s  is  ^  to  a  f  by  construction  .'.    k  e   and  f  s  are 

parallel. 

.*.  ak  =  ks,  (Ref.  IV)  If  a  line  be  drawn  through 
two  sides  of  a  A  parallel  to  the  third  side  it  divides  the 
sides  proportionally. 

.'.  a  k  =  ^  a  s. 

1  e  is  j^  to  and  bisects  a  k  at  r  . '.  is  parallel  to  s  t. 
.-.  a  r  =  ^  a  k  =  ^  a  s,  etc. 
.-.  a  1  =  ^  a  t  (Ref.  IV.) 
.-.  By  same  proof  a  o  =^  ^  a  X  etc. 
.-.  as  =  2ak;at  =  4al;  a  X  =8ao,  etc.,  in- 
definitely until  the  limit  is  reached. 

( Ref.  V.  The  difference  between  a  variable  and  its 
limit  is  a  variable  whose  limit  is  zero.) 

Then  a  j  =  arc  a  k  f . 
.-.  a  j  is  the  required  line. 

22 


PROBLEM  II 

To  draw  a  square  =  in  area  to  a  circle. 

Take  any  circle  a  f  c  g  Figure  3.  Draw  diameter 
a  c;  make  a  d  ^  J^  circumference  of  circle  as  per  pro- 
blem I.  Draw  d  b  j_  to  a  c  cutting  the  circumference 
at  b;  connect  a  b;  on  a  b  construct  the  square  a  b  i  h 
which  is  the  required  □• 

PROOF 

Post:  The  circumference  of  a  circle  X  ^  radius 
=  area. 

Post:  ^  the  circumference  of  a  circle  X  radius 
=  area. 

.*.  ^  circumference  X  2  radius  =  area. 

.*.  a  d  X  a  c  =  area  of  circle  a  f  c  g.  a  b"-^  =  a  d 
X  a  c.  (If  from  a  point  in  the  circumference  a  j^  be 
drawn  to  the  diameter,  and  chords  from  the  point  to 
the  extremities  of  the  diameter,  the  j_  is  a  mean  propor- 
tional between  the  segments  of  the  diameter  and  each 
chord  is  a  mean  proportional  between  its  adjacent  seg- 
ments and  the  dian>eter. ) 

.*.  a  d:  a  b  ::  a  b:  a  c. 
.-.  a  b-  =  a  d  X  a  c. 

28 


But  a  d  X  a  c  =  area  of  circle  a  f  c  g  as  above,  and 
a  b'^  =  area  of  square. 

. '.  the  square  a  b  i  h  :=  circle  a  f  c  g  which  was  to 
be  constructed. 


Note. — The  angle  bac  Figure  3,  when  construct- 
ed, may  be  used  for  solving  problems  2,  3,  4  and  5,  and 
other  allied  problems,  by  constructing  equal  angles. 
(The  nearest  approximate  to  angle,  b  a  c  is  27*^*36'  — 
and  for  convenience,  for  practical  work,  it  may  be 
marked  on  a  protractor,  or  a  triangle  containing  the 
essential  angle  constructed  of  wood  metal  or  other  ma- 
terial may  be  used.) 


PROBLEM  III 

To  draw  a  circle  =  in  area  to  a  square. 

Take  any  square  a  b  i  h,  Figure  3,  on  a  b  at  a  con- 
struct <Cbac=<bac,  Figure  3;  on  b  a  at  b  construct 
<abn  =  <^bac,  Figure  3;  with  e  as  center,  and  e 
a  or  e  b  as  radius  draw  circle  a  f  c  g;  then  the  area  of 
the  circle  a  f  c  g  =  area  of  the  square  a  b  i  h. 

24 


PROOF 

This  problem  is  the  converse  of  problem  II,  and  the 
reader  may  apply  the  proof. 

PROBLEM  IV 

To  draw  a  circle  with  circumference  =  a  straight 
line. 

Take  any  line  and  divide  in  into  4  =  parts;  let  a  d 
Figure  3,  represent  %  line;  at  d  erect  a  ^  d  b;  on  line 
a  d  at  a  construct  <bac=:<bac;  Figure  3,  the 
line  a  b  cutting  or  being  produced  until  it  cuts  d  b  at  b; 
draw  u  e  the  ]_  bisector  of  a  b  cuttino:  a  d  at  e;  with 
e  as  center,  and  e  a  or  e  b  as  radius,  draw  circle  a  f  c  g; 
then  circumference  of  circle  afcg=^4ad. 

PROOF 

The  proof  in  problem  II  may  be  applied  to  problem  IV. 

PROBLEM  V 

To  divide  any  angle  into  any  number  of  equal 
angles. 

25 


Let  b  a  c,  Figure  4,  be  the  given  angle  which  it 
is  proposed  to  divide  into  any  number  of  equal  parts, 
as  for  example,  five. 

Take  a  point  i  on  the  line  a  c;  make  a  i  =  1-2  ^ 
2-3  =  3-4  =  4-5;  with  a  as  center  and  a  i  as  radius 
describe  the  arc  i  d;  with  a  as  center  and  a  5  as  radius 
describe  the  arc  5  k  j ;  with  a  as  center  describe  a  num- 
ber ofindefinite  arcs,  as  e  e=  i  i,  etc.;  with  a  radius  = 
chord  I  d  mark  5  points  on  each  arc  5  k  j,  e  e,  i  i,  etc.; 
then 

5  chords  in  5  k  =  5  chord  i  d. 

5  chords  in  e  e  =  5  chord  i  d. 

5  chords  in   i  i  =  5  chord  i  d. 

Through  the  pointsi  e  k,  etc.,  draw  the  plane  curve 
line  i  e  f  k  cutting  the  line  b  a  at  f;  with  a  as  center 
and  a  f  as  radius  describe  the  arc  f  g  h;  make  chord  h 
g  ^  chord  I  d;  then  5  chords  in  f  h  =  5  chord  i  d, 
and  chord  h  g  =  chord  i  d  and  angle  g  a  h  =  ^ 
angle  b  a  c. 

PROOF 

5  chords  in  5  k  =  5  chord  i  d   by  construction. 
5       "        "   e  e  =  5       "id 
5       "        «'    i  i  =  5       •*     I  d 

26 


(( 


u  {« 


fh  =  5^  =  G6  =  iit)y  construction,  (by  chord 
-=  I  d.) 

.*.  f  h  =  5,  I  d  by  construction,  (by  chord  =  i  d.) 

Chord  h  g  =  chord  i  d  by  construction. 

5  chords  in  f  h  :=  5  chords  i  d  by  construction. 

h  g  =  i  f  h  by  construction. 

.*.  angle  g  a  c  =  ^-  angle  b  a  c. 

PROBLEM  VI 

To  draw  a  line  =  square  of  any  given  line. 

Take  any  line  a  b,  Figure  5;  on  a  b  or  on  a  b  pro- 
duced, take  a  definite  length  a  e  (not  less  than  j4  the 
given  line)  with  e  as  center  and  e  a  as  radius  describe 
the  semi-circle  a  f  c;  with  a  as  center  and  a  b  as  radius 
describe  the  arc  b  f  cutting  the  semi-circle  at  f;  from  f 
drop  the  }_fd;  then  ad  X  a  c  =  a  b-;  but  a  c  =  2  a 
e  and  a  e  is  definite  by  construction;  .'.  a  c  being  =  2 
a  e  is  definite  by  construction;  .*.  a  b  (which  is  indef- 
inite) taken  a  c  (definite by  construction)  times  equals 
the  square  of  a  b. 

PROOF 

a  f  ^  a  b  by  construction. 

But  af^  =  adXac(if  from  a  point  in  the  cir- 


cumference  a  j^  be  drawn  to  the  diameter,  and  chords 
from  the  point  to  the  extremities  of  the  diameter,  the 
_j_  is  a  mean  proportional  between  the  segments  of  the 
diameter  and  each  chord  is  a  mean  proportional  between 
its  adjacent  segments  and  the  diameter.) 

.*.  a  b-  =  a  d  X  a  c. 

.'.ad  taken  a  c  times  =  a  b  squared. 

PROBLEM  VII 

To  draw  a  line  =  any  given  line  X  by  any  other 
given  line. 

Take  an}"  two  lines  a  b  and  be,  Figure  6,  construct 
on  a  c  the  semi  circle  age  and  erect  the  x  &  b-  then  g 
b2  =  a  b  X  b  c. 

On  the  line  b  g  or  on  b  g  produced  take  a  definite 
length  b  f  (not  less  than  ,^1;  b  g  )  with  f  as  center  and 
f  b  as  radius  describe  the  semi-circle  b  i  h;  with  b  as 
center  and  b  g  as  radius  describe  the  arc  g  i  cutting  the 
semi-circle  b  i  h  at  i;  draw  i  d  j^  to  b  h;  then  i  h'^  =  h 
d  X  h  b  and  i  b-  =  b  d  X  b  h;  but  b  h  =  2  b  f  is  def- 
inite by  construction  .".  b  h  is  definite  by  construction; 
.•.a  line  may  be  drawn  =  d  b  taken  h  b  times  =  b  %^ 
(b  g  =  b  i)=  a  bXb  c.    (See  problem  VI  ,  Figure  5.) 

28 


PROOF 

b  g-^  =  a  b  X  b  c. 

(Ref.  I.  It  from  a  point  in  the  circumference  a  j_ 
be  drawn  to  the  diameter  and  chords  from  the  points  to 
the  extremities  of  the  diameter,  the  j_  is  a  mean  pro- 
portional between  its  adjacent  segment  and  the  diam- 
eter. ) 

A  line  may  be  drawn  =  square  of  any  line  as  per 
problem  VI. 

b  h  =  2  b  f  is  definite  by  construction. 

.'.  b  d  X  b  h  =  bg-  =  a  b  X  be  =  required  line. 

PROBLEM   VIII 

To  draw  a  line  =  cube  of  an}-  given  line. 

Take  any  line  a  e,  Figure  7;  on  the  line  a  e  or  on 
the  line  a  e  produced  take  a  definite  length  a  c  (not 
less  than  ^  the  given  line)  with  c  as  center  and  c  a  as 
radius  describe  the  semi-circle  a  f  b;  with  a  as  center 
and  a  e  as  radius  cut  the  arc  a  f  b  at  f;  draw  f  d  j^  a  e; 
then  a  d  X  a  b  =  a  e-  as  per  problem  VI. 

Draw  a  line  ^  a  e  X  a  e-  (a  line  may  be  drawn  = 
any  line  X  by  any  other  line,  problem  VII;)  then  a  d  X 
a  b  X  a  e  =  a  e'^ 

29 


PROOF 

a  b  =  2  a  c  is  definite  by  construction. 

.'.a  line  may  be  drawn  =^  a  d  X  a  b  X  a  e"-  as  per 
problem  6. 

a  e  =  the  given  line. 

.'.  a  line  =  a  e  X  a  e-  may  be  drawn  as  per  pro- 
blem 7. 

.*.  a  line  niav  be  drawn  ^  a  d  >    ab  X  ae  =  a  e'\ 


30 


YB 


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